If int x^3 e^{2 x} d x = frac{e^{2 x}}{8} f(x | vedclass

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(A) Given,$\int x^3 e^{2 x} d x = \frac{e^{2 x}}{8} f(x) + c$.Applying integration by parts repeatedly:$\int x^3 e^{2 x} d x = x^3 \frac{e^{2 x}}{2} -

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$\frac{1}{2}$

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(A) Given,$\int x^3 e^{2 x} d x = \frac{e^{2 x}}{8} f(x) + c$.Applying integration by parts repeatedly:$\int x^3 e^{2 x} d x = x^3 \frac{e^{2 x}}{2} - \int 3x^2 \frac{e^{2 x}}{2} d x$$= \frac{x^3 e^{2 x}}{2} - \frac{3}{2} \left( x^2 \frac{e^{2 x}}{2} - \int 2x \frac{e^{2 x}}{2} d x \right)$$= \frac{x^3 e^{2 x}}{2} - \frac{3x^2 e^{2 x}}{4} + \frac{3}{2} \int x e^{2 x} d x$$= \frac{x^3 e^{2 x}}{2} - \frac{3x^2 e^{2 x}}{4} + \frac{3}{2} \left( x \frac{e^{2 x}}{2} - \int \frac{e^{2 x}}{2} d x \right)$$= \frac{x^3 e^{2 x}}{2} - \frac{3x^2 e^{2 x}}{4} + \frac{3x e^{2 x}}{4} - \frac{3}{4} \frac{e^{2 x}}{2} + c$$= \frac{e^{2 x}}{8} (4x^3 - 6x^2 + 6x - 3) + c$.Comparing with the given form,$f(x) = 4x^3 - 6x^2 + 6x - 3$.For $f(x) = 1$,we have $4x^3 - 6x^2 + 6x - 4 = 0$,which simplifies to $2x^3 - 3x^2 + 3x - 2 = 0$.Factoring,$2(x^3 - 1) - 3x(x - 1) = 0 \Rightarrow (x - 1)(2(x^2 + x + 1) - 3x) = 0$.$(x - 1)(2x^2 - x + 2) = 0$.The roots are $x = 1$ (real) and the roots of $2x^2 - x + 2 = 0$ (complex).The sum of the complex roots is $-\frac{b}{a} = -(\frac{-1}{2}) = \frac{1}{2}$.

If $\int x^3 e^{2 x} d x = \frac{e^{2 x}}{8} f(x) + c$,then the sum of all the complex roots of $f(x) = 1$ is

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